Let $A$ be a $n\times n$ matrix with complex entries such that $A^m=I$ for some positive integer $m$. then
1) $A$ is diagonalisable matrix.
2) $A$ is similar to a triangular matrix but not $A$ need not be diagonalisable matrix.
3) all the eigenvalues of $A$ are roots of unity.
4) none of the above.
Since $m$ is some positive integer, always not prime. So my thinking says option 2 and 3 are correct. Is my answer correct?
The matrix $A$ satisfies $p(A) = 0$, where $p(x) = x^m - 1$. Because $p$ has no repeated factors, it must be the case that $A$ is diagonalizable (and each eigenevalue $\lambda$ of $A$ satisfies $p(\lambda) = 0$). So, 1 is correct.
Every matrix is similar to a triangular matrix, $A$ included. However, $A$ is necessarily diagonalisable. So 2 is incorrect as stated; I suspect that you have copied it incorrectly.
3 is correct. As I stated in the first paragraph, each eigenvalue $\lambda$ of $A$ satisfies $p(\lambda) = 0$, and hence $\lambda^m = 1$.